Question

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.5 m above the floor. A compressed-air cannon sits on the floor, 4.0 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.0 m above the floor. What was the launch speed of the plastic ball?

Short answer

The launch speed of the ball is$7.4\text{m/s}$.

Step-by-step solution

Step 1. Given information

The vertical height of the electromagnet is $3.5\text{m}$, the horizontal distance of the electromagnet from the canon is $4.0\text{m}$. The collision occurs at vertical height of $1.0\text{m}$above the floor. The masses of the plastic ball and the steel ball are $25\text{g}$and$35\text{g}$respectively.

Step2. Calculation

The formula to calculate the horizontal distance $\left(x\right)$travelled by the plastic ball is given by

$x=v_0\cos \theta t..................\left(1\right)$

Here, $v_0,\theta ,t$are the launch speed, the projection angle and the time taken by the plastic ball respectively.

Substitute $4.0\text{m}$for $x$into equation (1) to obtain the relation.

$v_0\cos \theta t=4..................\left(2\right)$

The formula to calculate the final vertical position $\left(y\right)$of the plastic ball is given by

localid="1648261941848" $\begin{array}{rcl}y& =& v_0\sin \theta t-\frac{1}{2}g{t}^2..........................\left(3\right)\end{array}$

Substitute $1.0\text{m}$for $y$, $3.5\text{m}$for localid="1648261988820" $v_0\sin \theta t$and $9.80{\text{m/s}}^2$for $g$into equation (5) and solve to calculate the time taken by the plastic ball to collide with the steel ball.

$\begin{array}{rcl}1.0\text{m}& =& 3.5\text{m}-\frac{1}{2}\times 9.80{\text{m/s}}^2\times t\\ t& =& \frac{3.5\text{m}-1.0\text{m}}{4.9{\text{m/s}}^2}\\ & \approx & 0.71\text{s}\end{array}$

Substitute $0.71\text{s}$for $t$into equation (2) and simplify to obtain the square of the cosine of the projection angle.

$\begin{array}{rcl}{v}_0\cos \theta \times 0.71& =& 4\\ \cos \theta & =& \frac{4}{0.71v_0}\\ {\cos }^2\theta & =& {\left(\frac{4}{0.71v_0}\right)}^2.....................\left(6\right)\end{array}$

Substitute $0.71\text{s}$for $t$and $3.5\text{m}$for $h$into equation (4) and simplify to obtain the square of the sine of the projection angle.

$\begin{array}{rcl}{v}_0\sin \theta \times 0.71& =& 3.5\\ \sin \theta & =& \frac{3.5}{0.71v_0}\\ {\sin }^2\theta & =& {\left(\frac{3.5}{0.71v_0}\right)}^2........................\left(7\right)\end{array}$

Add equation (6) and equation (7) and solve to calculate the required launch speed.

$\begin{array}{rcl}{\cos }^2\theta +{\sin }^2\theta & =& {\left(\frac{4}{0.71v_0}\right)}^2+{\left(\frac{3.5}{0.71v_0}\right)}^2\\ {v_0}^2& =& {\left(\frac{4}{0.71}\right)}^2+{\left(\frac{3.5}{0.71}\right)}^2\\ v_0& =& \sqrt{{\left(\frac{4}{0.71}\right)}^2+{\left(\frac{3.5}{0.71}\right)}^2}\\ & \approx & 7.4\end{array}$